What equation will create a 3D rose curve?

$\begingroup$

The parametric equation $x=a\cos(bt)\cos(t)$, $y=a\cos(bt)\sin(t)$ where $a$ & $b$ are constants and $t$ is parameter gives a rose curve which looks like,enter image description here

On a similar basis, is there a equation that gives a 3D rose curve? The curve would look like the surface formed by rotating each of the "petal" of the rose curve in 360 degrees along the radius vector.(I hope you get what I want to say -;)

$\endgroup$

2 Answers

$\begingroup$

This doesn't exactly answer the question, but with $k$, $m$, and $n$ positive integers, the parametric equations \begin{alignat*}{3} x(s, t) &= a\cos(mt) \cos^{k}(ns) &&\cos(t) &&\cos(s), \\ y(s, t) &= a\cos(mt) \cos^{k}(ns) &&\sin(t) &&\cos(s), \\ z(s, t) &= a\cos(mt) \cos^{k}(ns) &&\sin(s) && \end{alignat*} may provide some enjoyable plotting along similar lines.

For example, here's the surface with $m = 4$, $n = 1$, and $k = 8$:

A three-dimensional rose with eight lobes

The underlying idea is to take $\rho = \cos(m\theta)\cos^{k}(n\phi)$ in spherical coordinates $$ (x, y, z) = (\rho\cos\theta \cos\phi, \rho\sin\theta \cos\phi, \rho\sin\phi). $$

You may also enjoy learning about spherical harmonics.

$\endgroup$ 1 $\begingroup$

I made a video on this. Are you the one person that seen it and hit the like button on it?

I think these are nd rose curves. Assume x_value is from the spherical coordinates on the wikipedia page

r=x_1+x_2+..x_n


For 3d

$x_1=cos(\phi_1)*(x_1+x_2+x_3)$

$x_2=sin(\phi_1)*cos(\phi_2)*(x_1+x_2+x_3)$

$x_3=sin(\phi_1)*sin(\phi_2)*(x_1+x_2+x_3)$

$\endgroup$ 4

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

You Might Also Like